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Chapter 3: Problem 36

Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure \(3.1 .13,\) with an initial velocity \(v_{0}=300 \mathrm{ft} / \mathrm{s}\). The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. (a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by \(d^{2} s / d t^{2}=-g\) (equation (12) of Section 1.3 ). since \(d s / d t=v(t)\) the last differential equation is the same as \(d v / d t=-g,\) where we take \(g=32 \mathrm{ft} / \mathrm{s}^{2} .\) Find the velocity \(v(t)\) of the cannonball at time \(t\) (b) Use the result obtained in part (a) to determine the height \(s(t)\) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.

Short Answer

Expert verified
The maximum height without air resistance is approximately 1406.25 feet.

Step by step solution

01

Write down the differential equation for velocity

The equation describing the acceleration of the cannonball, ignoring air resistance, is given by \( \frac{dv}{dt} = -g \). Here, \( g = 32 \, \text{ft/s}^2 \). We will integrate this equation to solve for the velocity \( v(t) \).
02

Integrate to find velocity function

We integrate \( \frac{dv}{dt} = -32 \) with respect to \( t \) to obtain \( v(t) = -32t + C \). We use the initial condition \( v(0) = 300 \, \text{ft/s} \) to find the constant \( C \). Substituting \( t = 0 \) gives us \( 300 = C \), thus \( v(t) = -32t + 300 \).
03

Integrate velocity to find height

From the relation \( \frac{ds}{dt} = v(t) \), substitute \( v(t) = -32t + 300 \). Integrate \( \frac{ds}{dt} = -32t + 300 \) with respect to \( t \) to solve for \( s(t) \): \[ s(t) = \int (-32t + 300) \, dt = -16t^2 + 300t + C'. \]
04

Determine integration constant for height function

Using the initial condition that the height at \( t = 0 \) is zero, \( s(0) = 0 \), we find \( C' \). Substituting \( t = 0, s(0) = 0 \) gives \( 0 = 0 + C' \), so \( C' = 0 \). Therefore, \( s(t) = -16t^2 + 300t \).
05

Find the time when the cannonball reaches maximum height

The cannonball reaches its maximum height when its velocity is zero. Set \( v(t) = 0 \): \( -32t + 300 = 0 \). Solving for \( t \), we find \( t = \frac{300}{32} \approx 9.375 \) seconds.
06

Calculate maximum height using height function

Substitute \( t \approx 9.375 \) into the height function \( s(t) = -16t^2 + 300t \). This gives \( s(9.375) = -16(9.375)^2 + 300(9.375) \approx 1406.25 \). Therefore, the maximum height is approximately 1406.25 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projectile Motion
Projectile motion is a type of motion experienced by an object that is projected into the air and influenced only by gravity and its initial velocity. In this exercise, the motion is simplified as we ignore air resistance, focusing entirely on the forces of gravity acting upon the cannonball. This type of motion is characterized by a parabolic trajectory. As the cannonball moves upwards, gravity works against its initial velocity, eventually bringing it to a momentary halt, before accelerating it downwards. Understanding projectile motion helps us solve real-world problems involving any kind of objects thrown or dropped in a gravitational field.
Integration in Calculus
Integration is a fundamental concept in calculus used to find quantities like area, volume, and in this case, aspects of motion such as velocity and displacement. In the given exercise, integration is used twice; first to find the velocity function from the acceleration, and second to find the position function or height from the velocity.
  • The first integration transforms the acceleration, which is a constant \( a = -g \) into the velocity function \( v(t) = -gt + C\).
  • The constant \( C \) is determined using the initial velocity condition. Here, the result shows how the cannonball's velocity changes linearly over time, due to gravity's effect.
  • The second integration takes the velocity function to determine \( s(t) \), the position or height function. The integration process involves calculating the area under the velocity-time graph, showing how height changes as time passes.
These integrations demonstrate a classic application of calculus in physics.
Kinematics
Kinematics is the branch of physics that describes the motion of objects using equations and concepts without considering the forces causing the motion. In the exercise, we apply kinematic principles to model the cannonball's trajectory while ignoring air resistance. We use basic kinematics equations, derived from calculus, to understand both velocity and displacement.
  • The velocity equation \( v(t) = -32t + 300 \) shows how the velocity of the cannonball changes as time progresses.
  • Displacement or height is described by \( s(t) = -16t^2 + 300t \), a quadratic equation providing a time-height relationship, forming a parabolic path.
  • The maximum height is determined when the velocity equals zero, a common kinematic scenario highlighting the transition from upward to downward motion.
Kinematics in this exercise showcases how motion can be predicted mathematically.
Mathematical Modeling
Mathematical modeling involves using mathematical structures and equations to represent real-world scenarios. Here, we use differential equations to represent the motion of the cannonball. These equations offer a simplified model by neglecting air resistance, focusing on gravity and initial velocity.

Purpose of Mathematical Modeling:

  • To predict outcomes such as maximum height and velocity at any given time.
  • To provide a clear framework for analysis, allowing assumptions and simplifications.
  • To facilitate the use of calculus tools like differentiation and integration for deeper insights.
Mathematical models allow us to simulate, analyze, and understand complex situations with significant accuracy, backed by mathematical logic and calculations.

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Most popular questions from this chapter

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